A quadratic equation ax^2 + bx + c = 0 has roots that differ by the number 3. Let the roots be r and r + 3
We now have (x – r) (x – r – 3) = 0
=> x^2 – rx – x(r + 3) + r(r + 3) = 0
=> x^2 – x (r + r + 3) + r^2 + 3r = 0
=> x^2 – x (2r + 3) + r^2 + 3r = 0
Therefore a quadratic equation of the form x^2 – x (2r +3) + r^2 + 3r = 0, where r can be any number, has roots with a difference of 3.
Sice the roots of a quadratic equation differs by 3, we assume that one root is k and the other root is k+3.
Therefore the the required quadratic equation is (x-k)(x-k-3) = 0.
x^2-(k+k+3)x +k(k+3) = 0.
x^2-(2k+3)x+k(k+3) = 0 is the required equation.
Therefore for any chosen value of k, x^2-(2k+3)x+k(k+3) = 0 represents the equation with two roots , their (the roots') difference being 3.